Draw a sketch of the plane with coordinate axes $x$ and $y$, and mark on it the region described as $$\{(x,y) \colon |\min(x,y) | < 1\}.$$ Then think of probabilities, not before you have finished the sketch.
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Dilip SarwateNov 16 '11 at 20:08

I did. And I got a L-shape graph but I don't know how to start from here.
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geraldgreenNov 16 '11 at 20:09

Further hint on the L-shaped region. It can be partitioned into two rectangular regions and you can find the probability that $(X,Y)$ belongs in each, and add. The two probabilities are different. Or you can use express the region as $A\cup B$ where $P(A) = P(B)$ and use $P(A\cup B) = P(A) + P(B) - P(A\cap B)$.
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Dilip SarwateNov 16 '11 at 23:52

Your intuition is correct but you should be careful about how you write up the solution, since the areas of the regions are not finite. Write the same thing in terms of the probabilities (and add another line to go from $P\{-1 < X < 1, Y > 1\}$ to $P\{-1 < X < 1, Y < -1\}$ so that $$P\{-1 < X < 1, Y > -1\}+ P\{-1 < X < 1, Y < -1\} = P\{-1 < X < 1\}$$ is immediately obvious, and I will be glad to upvote your answer.
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Dilip SarwateNov 17 '11 at 2:07